Find the direction angles of the vector. Round to the nearest degree, if necessary.
-$$\mathbf { v } = \mathbf { i } - \mathbf { j } + 2 \mathbf { k }$$
A) $\alpha = 66 ^ { \circ } , \beta = 114 ^ { \circ } , \gamma = 35 ^ { \circ }$
B) $\alpha = 80 ^ { \circ } , \beta = 100 ^ { \circ } , \gamma = 71 ^ { \circ }$
C) $\alpha = 63 ^ { \circ } , \beta = 117 ^ { \circ } , \gamma = 26 ^ { \circ }$
D) $\alpha = 69 ^ { \circ } , \beta = 111 ^ { \circ } , \gamma = 45 ^ { \circ }$

Question

Quizplus · Accepted Answer

The direction angles $\alpha$, $\beta$, and $\gamma$ of a vector $\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$ can be found using the dot product and the magnitude of the vector. The direction angles are given by $\cos^{-1}\left(\frac{a}{|\mathbf{v}|}\right)$, $\cos^{-1}\left(\frac{b}{|\mathbf{v}|}\right)$, and $\cos^{-1}\left(\frac{c}{|\mathbf{v}|}\right)$ respectively. For $\mathbf{v} = \mathbf{i} - \mathbf{j} + 2\mathbf{k}$, the magnitude $|\mathbf{v}|$ is $\sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{6}$. Thus, $\alpha = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right) \approx 66^\circ$, $\beta = \cos^{-1}\left(\frac{-1}{\sqrt{6}}\right) \approx 114^\circ$, and $\gamma = \cos^{-1}\left(\frac{2}{\sqrt{6}}\right) \approx 35^\circ$.

Find the Direction Angles of the Vector v=i−j+2k\mathbf { v } = \mathbf { i } - \mathbf { j } + 2 \mathbf { k }v=i−j+2k

Find the Direction Angles of the Vector $\mathbf { v } = \mathbf { i } - \mathbf { j } + 2 \mathbf { k }$